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14.2.5

Example: Static analysis of a truss with roof load

W B

The figure to the right shows a roof load of weight W that is supported at point B of a planar truss.a The truss is attached to ground at point A by a pin-joint and point C by a smooth (frictionless) pin-roller joint. The roof load and truss are in static equilibrium.

a A truss has two-force members and is considered massless relative to the loads it carries. As shown by Homework 12.??, forces applied to the end of each member must be parallel to the line connecting the member's end-points.

A

L L

C

The steps for analyzing this structure are:1 1. Identify a system S. 2. Identify the relevant external contact and distance forces on S (or their equivalent force systems). 3. Introduce points, unit vectors, scalar symbols (e.g., force or torque measures), etc, (as necessary).

Note: Use Newton's law of action/reaction to reduce the introduction of unknown force/torque measures.

4. Use laws of static or dynamic equilibrium, e.g, for statics: FS = 0 and/or MS/P = 0. Form scalar equations via vector dot-products. 5. If there are an insufficient number of equations for a determinate system to answer the questions "what are the forces" and "where is it", identify another system S and repeat. The 2nd column of the following table shows the magnitude of the force in each member as a function of L, and W . The 3rd column shows whether each member is in compression (forces on the member try to shorten it) or tension (forces on the member try to elongate it) when W is positive. The 4th column shows the minimum and maximum load in each member for values of between 0 and 90 .

Note: Forces in member BC can be determined by symmetry.

Member AB AC BC

Force magnitude

W 2 sin() W cos() 2 sin() W 2 sin()

Compression or tension Compression Tension Compression

Minimum load W 2 at = 90 0 at = 90 W 2 at = 90

Maximum load at = 0 at = 0 at = 0

One incorrect conclusion from this analysis is 90 is structurally superior to smaller values of . Since members AB and BC are in compression and a member's chance of buckling increases significantly as its length increases,2 one must trade-off decreasing axial compressive loads with an increased chance of buckling. When buckling dominates a design, additional members (sometimes carrying no-load) are added near the center of each member to decrease the possibility of buckling.

The Autolev analysis for this truss structure is provided in Section 14.4. Euler's critical buckling load Fbuckling for a long simply-supported column under an axial compressive force is Fbuckling = c where Lcolumn is the length of the column (member) and c is a constant that depends on the column's elastic modulus 2 Lcolumn and area moment of inertia.

2

1

Copyright c 1992-2009 by Paul Mitiguy

143

Chapter 14:

Moments and torque

14.4

Static analysis of the truss of Section 14.2.5 with Autolev

W B

(1) % File: TrussABCTopLoad.al (Truss analysis) (2) %-------------------------------------------------------------------(3) NewtonianFrame N % Ground (4) RigidFrame S % Entire truss (5) Point A(S), B(S), C(S) % Nodes on truss (6) %-------------------------------------------------------------------(7) % Constants and variables (8) Constant L % Twice the distance between A and C (9) Constant theta % Angle between AC and AB A (10) Constant W % Weight applied to node B (11) Variable FAx % Nx> measure of external force on A (12) Variable FAy % Ny> measure of external force on A (13) Variable FCy % Ny> measure of external force on E (14) Variable FAB % Force on A from AB member directed from A to B (15) Variable FAC % Force on A from AC member directed from A to C (16) %-------------------------------------------------------------------(17) % Relevant external contact and distance forces on S (18) A.AddForce( FAx*Nx> + FAy*Ny> ) -> (19) Force_A> = FAx*Nx> + FAy*Ny> (20) B.AddForce( -W*Ny> ) -> (21) Force_B> = -W*Ny> (22) C.AddForce( FCy*Ny> ) -> (23) Force_C> = FCy*Ny> (24) (25) (26) -> (27) %-------------------------------------------------------------------% Static analysis of entire system ResultantForceOnS> = S.GetResultantForce() ResultantForceOnS> = FAx*Nx> + (FAy+FCy-W)*Ny>

L L

C

(28) MomentOfSAboutA> = Cross( L*Nx>, -W*Ny> ) + Cross( 2*L*Nx>, FCy*Ny> ) -> (29) MomentOfSAboutA> = -L*(W-2*FCy)*Nz> (30) ZeroSystem[1] = Dot( ResultantForceOnS>, Nx> ) -> (31) ZeroSystem[1] = FAx (32) ZeroSystem[2] = Dot( ResultantForceOnS>, Ny> ) -> (33) ZeroSystem[2] = FAy + FCy - W (34) ZeroSystem[3] = Dot( MomentOfSAboutA>, -> (35) ZeroSystem[3] = -L*(W-2*FCy) (36) Solve( ZeroSystem, FAx, FAy, FCy ) -> (37) FAx = 0 -> (38) FAy = 0.5*W -> (39) FCy = 0.5*W (40) (41) (42) -> (43) %-------------------------------------------------------------------% Relevant external contact and distance forces on pin A UnitVectorFromAToB> = Vector( N, cos(theta), sin(theta), 0 ) UnitVectorFromAToB> = cos(theta)*Nx> + sin(theta)*Ny> Nz> )

(44) A.AddForce( B, FAB * UnitVectorFromAToB> ) -> (45) Force_A_B> = cos(theta)*FAB*Nx> + sin(theta)*FAB*Ny> (46) A.AddForce( C, FAC * Nx> ) -> (47) Force_A_C> = FAC*Nx> (48) (49) (50) -> (51) %-------------------------------------------------------------------% Static analysis of node A ResultantForceOnA> = A.GetResultantForce() ResultantForceOnA> = (FAC+FAx+cos(theta)*FAB)*Nx> + (FAy+sin(theta)*FAB )*Ny>

(52) ZeroA = Dot( ResultantForceOnA>, [Nx>; Ny>] ) % Creates 2x1 matrix -> (53) ZeroA = [FAC + FAx + cos(theta)*FAB; FAy + sin(theta)*FAB] (54) (55) -> (56) -> (57) AutoRhs ALL % Ensure results are explicit in W, L, and theta Solve( ZeroA, FAB, FAC ) FAB = -0.5*W/sin(theta) FAC = 0.5*W*cos(theta)/sin(theta)

Copyright c 1992-2009 by Paul Mitiguy

146

Chapter 14:

Moments and torque

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